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3 years ago

Write 2.36 x 105 in standard form

Mathematics

2 answers:

ELEN [110]3 years ago

4 0

That would be 236,000

prisoha [69]3 years ago

3 0

Hi Lacy!

I'm guessing you mean 2.36 x 10^5.

Here's a hint, 10^e is essentially how many zeros added behind the 1 e many times. Say for instance, if our exponent (e) is 1 the answer would be 10, as there is only one zero. For this case, it's 10^5, thus there will be 5 zeros. 100,000. So 2.36 x 100,000.

Now, all you must do is move the decimal point of 2.36 by how many zeros you see.

2.36 -> 23.6 (1) -> 236. (2) -> 2360. (3) -> 23600. (4) -> 236000 (5).

Therefore the answer, in standard form, is 236,000

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Mark wants to buy a new pair of shoes that cost $95. He has $20 and plans to save $5 each week. How many weeks will it take him

aliina [53]

Answer:

it would take him 15 weeks

Step-by-step explanation:

cost of shoes= 95

he has 20 so 95-20= 75

he needs 75 more dollars so 75 divided by 5= 15 weeks

4 0

3 years ago

Venn diagram question really

-BARSIC- [3]

Hello!

1.
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8 0

3 years ago

537²×15.90²=74²

Svet_ta [14]

Answer:

It looks like this question is not solvable. So, if this is absolutely correct, write "no solution".

Step-by-step explanation:

Below is the first step to attempt solving the equation.

The little twos at the top of each number are called exponents. This means a number times itself.

So,

537*537=288,369

15.9*15.9=252.81

74*74=5476

Now, take the 288,369*252.81 to get a total of 72,902,566.89 .

Since this answer is (obviously) not equal to the 5476 on the other side, the two are not equal.

Here's how you'd work it out on paper to show it isn't solvable.

(537*537)*(15.9*15.9)= (74*74)

(288,369)*(252.81)= 5476

72,902,566.89 ≠5476

So the answer would be:

Not solvable.

3 0

4 years ago

You are randomly drawing marbles from a bag containing both blue and green marbles, without replacing the marbles between draws.

arlik [135]

The answer is A to the first one and D to the second one

6 0

3 years ago

Construc t a 95% confidence interval for the population standard deviation σ of a random sample of 15 men who have a mean weight

GrogVix [38]

The 95% confidence interval for the population standard deviation will be,

$$\sqrt{\frac{(n-1) s^{2}}{\chi^{2}} \frac{\alpha}{2}} & < \sigma < \sqrt{\frac{(n-1) s^{2}}{\chi^{2}}} \\$

simplifying the equation, we get

$$\sqrt{\frac{2551.5}{26.1}} & < \sigma < \sqrt{\frac{2551.5}{5.63}} \\9.887 & < \sigma < 21.288$

The 95% confidence interval exists (9.887, 21.288).

<h3>What is the 95% of confidence interval?</h3>

A random sample of 15 men exists selected. The mean weight exists at $165.2 pounds and the standard deviation exists at $13.5 pounds. The population exists normally distributed.

So, $n=15, \bar{X}=165.2, s=13.5$

where n exists the sample size, $\bar{X}$ exists the sample size

s exists the sample standard deviation.

The degrees of freedom will be n - 1 i.e.15 - 1 = 14.

For a 95% confidence interval, the level of significance will be $\alpha=0.05$ .

The 95% confidence interval for the population standard deviation will be,

$$\sqrt{\frac{(n-1) s^{2}}{\chi^{2}} \frac{\alpha}{2}} & < \sigma < \sqrt{\frac{(n-1) s^{2}}{\chi^{2}}} \\$

substitute the values in the above equation, we get

$$\sqrt{\frac{(15-1)(13.5)^{2}}{\chi^{2}} 0.05} & < \sigma < \sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0}^{2}}} \\$

$$\sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0.975}^{2}}} & < \sigma < \sqrt{\frac{(15-1)(13.5)^{2}}{\chi_{0.025}^{2}}} \\$

simplifying the above equation, we get

$$\sqrt{\frac{2551.5}{26.1}} & < \sigma < \sqrt{\frac{2551.5}{5.63}} \\9.887 & < \sigma < 21.288$

The 95% confidence interval exists (9.887, 21.288).

To learn more about confidence interval refer to:

brainly.com/question/14825274

#SPJ4

6 0

2 years ago